Simplify; express your answer in exponential form. Assume $r\neq 0, k\neq 0$. $\dfrac{{(r^{-3})^{-1}}}{{(r^{-1}k^{3})^{2}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-3}}$ to the exponent ${-1}$ . Now ${-3 \times -1 = 3}$ , so ${(r^{-3})^{-1} = r^{3}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-1}k^{3})^{2} = (r^{-1})^{2}(k^{3})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-3})^{-1}}}{{(r^{-1}k^{3})^{2}}} = \dfrac{{r^{3}}}{{r^{-2}k^{6}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{3}}}{{r^{-2}k^{6}}} = \dfrac{{r^{3}}}{{r^{-2}}} \cdot \dfrac{{1}}{{k^{6}}} = r^{{3} - {(-2)}} \cdot k^{- {6}} = r^{5}k^{-6}$.